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I want to prove that $x = 0.5$ is the global maximum of the function

$f(x) = \frac{(1-a)^2e^{(2x\cdot(x-1)a^2)}}{(1-a)(e^{(2x\cdot(2x-1)a^2)}+e^{((2x-1)\cdot(2x-2)a^2)})-2(1-2a)e^{(4x\cdot(x-1)a^2)}}$

where $a\in(0,1)$ and $x\in[0,1]$. I tried to show this using "conventional" ways, but unfortunately the attempts have failed.

Thanks a lot!

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up vote 0 down vote accepted

I have to run out now, so will write only the hint as of now. Later I might complete the details. Let me do one case here, the other one probably follows similarly.

Let $a \in (1/2,1)$, and write $a=1/2 + \epsilon$, for an appropriate $\epsilon$.

Now consider the function $1/f(x)$, which is (after some cleanup) seen to be

\begin{equation*} \frac{2 e^{-2 (-1+x) x \left(\frac{1}{2}+\epsilon \right)^2} \left(e^{2 (-1+x) (-1+2 x) \left(\frac{1}{2}+\epsilon \right)^2} (1-2 \epsilon )+e^{2 x (-1+2 x) \left(\frac{1}{2}+\epsilon \right)^2} (1-2 \epsilon )+8 e^{(-1+x) x (1+2 \epsilon )^2} \epsilon \right)}{(1-2 \epsilon )^2} \end{equation*}

But this is a nonnegative (since $2\epsilon < 1$) combination of convex functions of $x$ (because we have exponentials, products are ok), and is thus itself convex. Now, you can numerically or directly optimize this function and see that its minimum occurs at $x=0.5$, which implies that the maximum of your function occurs at $x=0.5$. (assuming it is easy to guarantee that $f(x)$ is a positive function).

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Thank you for your respond!. But, I think that there is a little mistake. The term:$e^{-2(\frac{1}{2}+\epsilon)^2x(x-1)}$is not convex, and hence the product is also not surely convex. – Josh Jul 22 '12 at 9:05
By the way, you can assume that $f$ is non-negative. – Josh Jul 22 '12 at 9:08
Ok, i tried to open the product and "luckily" the product is convex! For $\alpha\in(0,\frac{1}{2})$ we can use the symmetry of $f$ around $\frac{1}{2}$. – Josh Jul 22 '12 at 9:56
Actually, this function is a speceific case of the function: $f(x) = \frac{(\int_0^{1-a}\exp(x\cdot(x-1)\cdot(g(x+a)-g(x))^2)dx)^2}{\int_0^{1-a}(\exp‌​(2x\cdot(2x-1)\cdot(g(x+a)-g(x))^2)+\exp((2x-1)\cdot(2x-2)\cdot(g(x+a)-g(x))^2))d‌​x-2\int_a^{1-a}\exp(x\cdot(x-1)\cdot(g(x+a)-g(x-a))^2)dx}$ where $g$ is some continues function. This function again has a globel maximum at $x = \frac{1}{2}$. If we substitute $f(x) = x$, it can be seen that we get the original function. Do you have any idea? Thanks again – Josh Jul 22 '12 at 9:57
@MarkT1: I think the general case might also yield to some nice "variable substitutions" but it will require some playing around with the formula. But it seems more likely that your general case may succumb better to arithmetic-geometric mean inequalities for bounding the solution. Another lemma that might be useful is: if $g$, $h$ are convex and both nondecreasing (or nonincreasing) and positive on an interval, then the product $gh$ is also convex. This lemma holds only on convex functions on $R$. – Suvrit Jul 22 '12 at 19:00

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